Technical Mathematics Grade 12 P1 NW June 2025 QP (Afr)

PROVINCIAL ASSESSMENT/ PROVINSIALE ASSESSERING GRADE/GRAAD 12 12 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 JUNE/JUNIE 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 MARKING CODES/NASIENKODES: A Accuracy/Akkuraatheid AO Answers only/Slegs antwoorde CA Consistent accuracy/Volgehoue akkuraatheid M Method/Metode R Rounding/Afronding NPR No penalty for rounding/Geen penalisering vir afronding nie NPU No penalty for units omitted/Geen penalisering vir eenhede weggelaat nie S Simplification/Vereenvoudig SF Substitution in correct formula/Vervanging in korrekte formule F Correct formula/Korrekte formule These marking guidelines consist of 13 pages. Hierdie nasienriglyne bestaan uit 13 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 2 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne NOTE:  If a candidate answers a question TWICE, only mark the FIRST attempt.  Consistent accuracy marking applies to all aspects of the marking guidelines where applicable as indicated with the marking code CA.  If a candidate has crossed out (cancelled) an attempt to a question and NOT redone the solution, mark the crossed out (cancelled) version. LET WEL:  Indien ‘n kandidaat die vraag TWEE keer beantwoord, sien slegs die EERSTE poging na.  Volgehoue akkuraatheid word toegepas in alle aspekte van hierdie nasienriglyne, waar nodig is dit aangedui met die merkkode CA.  As ‘n kandidaat ‘n antwoord van ‘n vraag doodtrek (kanselleer) en nie oordoen nie, sien die doodgetrekte (gekanselleerde) poging na. QUESTION/VRAAG 1 1.1.1 (2 − 6𝑥)3𝑥 = 0 1 2 − 6𝑥 = 0 or/of 3𝑥 = 0  𝑥=3 A 2 = 6𝑥 or/of 𝑥 = 0 1  𝑥=0 A = 𝑥 or/of 𝑥 = 0 3 AO: full marks/ volpunte (2) 1.1.2 2 = 3𝑥 − 4 𝑥 2 = 3𝑥 2 − 4𝑥  standard form/ 3𝑥 2 − 4𝑥 − 2 = 0 standaard vorm A −𝑏±√𝑏 2 −4𝑎𝑐 𝑥=  SF CA 2𝑎 −(−4)±√(−4)2 −4(3)(−2)  𝑥 = −0,39 𝑥=  𝑥 = 1,72 CA 2(3) 4±2√10 NPR 𝑥= 6 𝑥 = −0,39 or/of 𝑥 = 1,72 (4) 1.1.3 2 𝑥 + 3𝑥 ≥ 0  factors /faktore 𝑥(𝑥 + 3) ≥ 0 A 𝑥 = 0 or/of 𝑥 = −3  both 𝑥-values/beide x- waardes OR /OF CA −(3)±√(3)2 −4(1)(0) 𝑥= 2(1)  number line/ 𝑥 = 0 or/of 𝑥 = −3 getallelyn CA  end points and notation/eindpunte en I I I I I I notasie CA -3 0 (4) 𝑥 ≤ −3 or /of 𝑥≥0 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 3 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 1.2 𝑠 − 𝑞 = 1 …(1)  𝑠 𝑜𝑟/𝑜𝑓 𝑞 subject of 𝑠 2 + 𝑞 2 = 5 …(2) the formula/ 𝑠 = 𝑞 + 1 …(3) onderwerp van die formule (𝑞 + 1)2 + 𝑞 2 = 5 A 𝑞 2 + 2𝑞 + 1 + 𝑞 2 − 5 = 0  SF CA 2𝑞 2 + 2𝑞 − 4 = 0 𝑞2 + 𝑞 − 2 = 0  standard form/ (𝑞 + 2)(𝑞 − 1) = 0 standaard vorm 𝑞 = −2 or 𝑞 = 1 CA 𝑠 = (−2) + 1 or/of 𝑠 = (1) + 1 𝑠 = −1 or/of 𝑠 = 2  factors/faktore CA OR/OF 𝑠 − 𝑞 = 1 …(1)  both 𝑠-values/beide 𝑠 2 + 𝑞 2 = 5 …(2) s-waardes CA 𝑠 = 𝑞 + 1 …(3)  both 𝑞-values/beide 2 (𝑞 + 1) + 𝑞 = 5 2 q-waardes CA 𝑞 2 + 2𝑞 + 1 + 𝑞 2 − 5 = 0 2𝑞 2 + 2𝑞 − 4 = 0 −(2)±√(2)2 −4(2)(−4) 𝑞= 2(2) 𝑞 = −2 or/of 𝑞=1 𝑠 = (−2) + 1 𝑠 = (1) + 1 𝑠 = −1 𝑠=2 (6) 1.3 𝐴 = 𝑃(1 + 𝑖)𝑛 1.3.1 𝐴  dividing with 𝑃/deel = (1 + 𝑖)𝑛 𝑃 𝐴 met 𝑃 A log 𝑃 = log(1 + 𝑖)𝑛  Log property/Log 𝐴 log 𝑃 = 𝑛 log(1 + 𝑖) eienskap A 𝐴 log( )  answer/antwoord CA 𝑃 =𝑛 OR/OF log(1+𝑖)  dividing with 𝑃/Deel OR/OF met 𝑃 A change exponential 𝐴 to logarithm/verander = (1 + 𝑖)𝑛 eksponent na 𝑃 𝐴 𝑛 = log (1+𝑖) (𝑃) logaritmiese vorm CA AO: full marks/volpunte (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 4 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 1.3.2 log( 𝑅50000 )  SF 𝑅25000 𝑛= CA from/vanaf 1.3.1 log(1+8.5%) 𝑛 = 8,5 years/jaar  answer/antwoord CA OR/OF  SF CA 50000  answer/antwoord 𝑛 = log (1+8,5%) (25000) CA 8,5 years/jaar PR: 1 mark for incorrect rounding/1 punt vir foutiewe afronding (2) 1.4 𝐴 = 110012 and/en 𝐵 = 1011112 1.4.1 𝐵−𝐴 1011112  M A −110012  answer/antwoord = 101102 CA OR/OF 26 25 24 23 22 21 20 64 32 16 8 4 2 1  M CA 1 0 1 1 1 12 − 1 1 0 0 12  answer/antwoord = 1 0 1 1 02 CA AO: full marks/volpunte (2) 1.4.2 25 24 23 22 21 20  M CA 32 16 8 4 2 1 from/vanaf 1.4.1 1 0 1 1 02 101102 = 16 + 4 + 2  answer/ antwoord = 22 CA OR/OF 1011112 − 110012 AO: full marks/ = 47 − 25 volpunte = 22 (2) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 5 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 𝑥= −2±4√𝑘+3 3 2.1.1 𝑘 = −1 ∆= 𝑘 + 3 = (−1) + 3 =2 ∴ ∆> 0 the roots are real, unequal and irrational/  answer/ die wortels is reël, ongelyk en irrasionaal. antwoord A (1) 2.1.2 𝑘 = −3 ∆= 0 the roots are real, equal and rational/  answer/ die wortels is reël, gelyk en rasionaal. antwoord A (1) 2.1.3 𝑘 < −3  answer/ ∆< 0 the roots are non-real or imaginary/ antwoord A die wortels is nie-reël of imaginêr. (1) 2.2 𝑥 2 − 𝑥 = −2 − 𝑞 𝑥2 − 𝑥 + 2 + 𝑞 = 0  standard form/ ∆> 0 standaard 𝑏 2 − 4𝑎𝑐 > 0 vorm A (−1)2 − 4(1)(2 + 𝑞) > 0  ∆> 0 A −7 − 4𝑞 > 0  SF CA −7 > 4𝑞  answer/ 7 antwoord CA −4 > 𝑞 (4) [7] QUESTION/VRAAG 3 3.1.1 √3(√27 + 2√3) = √3(√9 × √3 + 2√3)  surds in brackets/ wortels in hakies = √3(3√3 + 2√3) A = 15  answer/antwoord CA Penalty/Penaliseer: 1 mark for AO/1 punt vir AO (2) 3.1.2 33𝑥−2 33𝑥+1 . 9𝑥−3  prime base/ priemfaktore A 33𝑥−2  product/produk = 3𝑥+1 3 . 32(𝑥−3) /quotient rule/kwosiënt reël A 33𝑥−2 = 5𝑥−5 3 = 33𝑥−5𝑥−2−(−5)  answer/antwoord 33 27 CA = 33−2𝑥 OR/OF = OR/OF = 2𝑥 32𝑥 3 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 6 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 3.1.3 log 5 +log 125 log 625− log 25  prime base/ log 5+log 53 priemfaktore A = log 54 −log 52  log property/log log 5+3log 5 eienskap CA = 4log 5−2log 5  factor/gemeenskaplike log 5(1+3) faktor CA = log 5(4−2)  answer/antwoord A =2 (4) 64×8 3.2 log 𝑥 64 + log 𝑥 8 − log 𝑥 32 = log 5 625  log 𝑥 A 32 64×8 log 𝑥 32 = log 5 54  log 5 54 A log 𝑥 16 = 4log 5 5  simplification/ vereenvoudiging CA log 𝑥 16 = 4(1)  exponential/ 4 𝑥 = 16 eksponensieël CA 𝑥 4 = 24  answer/antwoord 𝑥=2 CA AO: 1 mark/punt (5) 3.3 𝑧 = −√2 + 3𝑖 3.3.1 𝑧̅ = −√2 − 3𝑖  conjugate/ gekonjugeerde A (1) 3.3.2 Imaginary/Imaginêr (𝑖)  correct quadrants/ korrekte kwadrante CA  correct terminal arm/ Real/Reël korrekte terminaalarm −√2 CA (−√2; −3) -3 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 7 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 3.3.3 |𝑟| = √𝑎2 + 𝑏 2  SF modulus |𝑟| = √(−√2)2 + (3)2  |𝑟| = 2√2 CA |𝑟| = √11 𝑏 𝜃𝐴𝑟𝑔 = 180° + 𝑡𝑎𝑛−1 (𝑎)  SF ∠𝑟𝑒𝑓 CA −1 3 𝜃𝐴𝑟𝑔 = 180° + 𝑡𝑎𝑛 ( ) √2 𝜃𝐴𝑟𝑔 = 180° + 64,76059817 … °  argument CA 𝜃𝐴𝑟𝑔 = 244.76° NPR 𝑧 = 𝑟(cos 𝜃 + 𝑖 ∙ sin 𝜃)  polar form/polêre vorm 𝑧 = √11(cos(244,76°) + 𝑖 ∙ sin(244,76°)) CA (5) 3.4 𝑎 + 𝑏𝑖 = 2(3 − 2𝑖) − (−5𝑖)  simplification/ 𝑎 + 𝑏𝑖 = 6 − 4𝑖 + 5𝑖 vereenvoudiging A 𝑎 + 𝑏𝑖 = 6 + 𝑖  𝑎=6 CA 𝑎=6 and/en 𝑏=1  𝑏=1 CA (3) [25] QUESTION/VRAAG 4 4.1 𝑘(𝑥) = −𝑥 2 + 2𝑥 + 15 3 𝑞(𝑥) = 𝑥 − 4 4.1.1 𝑥=0  𝑥=0 A 𝑦 = −4  𝑦 = −4 A (2) 4.1.2 𝑥-intercept/ x-afsnit, 𝑦 = 0  𝑞(𝑥) = 0 A 3 0=𝑥−4 4𝑥 = 3 3 3  (4 ; 0) CA 𝑥=4 (2) 4.1.3 −𝑥 2 + 2𝑥 + 15=0  factors/faktore A 𝑥 2 − 2𝑥 − 15 = 0  (−3; 0) CA (𝑥 + 3)(𝑥 − 5) = 0  (5; 0) CA 𝑥 = −3 or/of 𝑥 = 5 (−3; 0) or/of (5; 0) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 8 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 4.1.4 𝑥 = − 𝑏  F axis of symmetry/ 2𝑎 simmetrie-as A (2) =−  SF A 2(−1) =1 4𝑎𝑐 − 𝑏 2 𝑦= 4𝑎 4(−1)(15) − (2)2 = 4(−1)  turning point/ = 16 draaipunt CA TP(1; 16) OR/OF 𝐾(1) = −(1)2 + 2(1) + 15 = 16 TP(1; 16) (3) 4.1.5 𝑘(𝑥)  intercepts/afsnitte CA  turning point/ draaipunt CA  shape/vorm CA 𝑞(𝑥)  asymptotes/ asimptote CA  intercept/afsnitte CA  shape/vorm CA (6) 4.2 𝑓(𝑥) = −√𝑟 2 − 𝑥 2 𝑞(𝑥) = 2𝑥 − 10 4.2.1 𝑥2 + 𝑦2 = 𝑟2  𝑔(𝑥) = 0 CA 𝐴(3; −4) (3)2 + (−4)2 = 𝑟 2  equation/ 25 = 𝑟 2 vergelyking CA 𝑦 = −√25 − 𝑥 2 (2) 4.2.2 𝐵(0; −5)  answer/antwoord CA (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 9 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 4.2.3 Domain/Definisieversameling:  end points/ eindpunte : −5 ≤ 𝑥 ≤ 5 OR/OF 𝑥 ∈ [−5; 5] CA  notation/notasie Range/ Waardeversameling:  end points/ eindpunte −5 ≤ 𝑦 < 0 OR/OF 𝑦 ∈ [−5; 0) CA  notation/notasie (4) 4.2.4 3 ≤ 𝑥 ≤ 5 OR/OF 𝑥 ∈ [3; 5]  end points/ eindpunte CA  notation/notasie (2) [25] QUESTION/VRAAG 5 5.1 𝑖 𝑖 = (1 + )𝑚 − 1  formula/formule A 𝑒𝑓𝑓 𝑚 11,5% 1×52  SF A 𝑖𝑒𝑓𝑓 = (1 + 52 ) −1 𝑖𝑒𝑓𝑓 = 0,12173099529 × 100  answer/antwoord 𝑖𝑒𝑓𝑓 = 12,17% CA (3) 5.2 𝐴 = 𝑃(1 − 𝑖𝑛)  formula/formule A 1 𝑃 = 𝑃(1 − 0,15𝑛)  SF 2 CA 0,5 = 1 − 0,15𝑛  simplification/ 0,15𝑛 = 1 − 0,5 0,5 vereenvoudiging 𝑛 = 0,15 CA 𝑛 = 3,33 years/jaar  answer/antwoord CA (4) 5.3 𝐴 = 𝑃(1 + 𝑖)𝑛  formula/formule A 7,5% 2,5×12  SF 𝐴 = 𝑅25 000(1 + )  𝑅30 138,1652589 12 𝐴 = 𝑅30 138,1652589 CA 8% 𝐴 = 𝑅30 138,1652589(1 + 4 )4×0,5 8% 𝐴 = 𝑅31 355,7471354 + 𝑅7500  (1 + 4 )4×0,5 8% 𝐴 = 𝑅38 855,7471354(1 + 4 )2×4 8% 𝐴 = 𝑅45 525,7006369959  (1 + 4 )4×2 𝐴 = 𝑅45 525,70  𝐴 = 𝑅45 525,70 CA Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 10 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne OR/OF OR/OF  formula/formule 𝐴 = 𝑃(1 + 𝑖)𝑛 A 7,5% 2,5×12 8% 2,5×4 7,5% 𝐴 = 𝑅25 000(1 + ) × (1 + )  𝑅25 000(1 + 12 )2,5×12 12 4 8% 8% 2×4  (1 + 4 )2,5×4 + 𝑅7 500 (1 + ) 4  +𝑅7500 𝐴 = 𝑅45 525,7006369959 8%  (1 + 4 )2×4  A=R45 525,70 CA 𝐴 = 𝑅45 525,70 NPR (6) [13] QUESTION/VRAAG 6 6.1 2 𝑓(𝑥) = 5 − 3 𝑥 𝑓(𝑥+ℎ)−𝑓(𝑥)  formula/formule A 𝑓 ′ (𝑥) = lim ℎ→0 ℎ 2 2 5− (𝑥+ℎ)−(5− 𝑥) 3 3  SF CA 𝑓′(𝑥) = lim ℎ→0 ℎ  simplification/ 2 2 2 5− 𝑥− ℎ−5+ 𝑥 vereenvoudiging CA 𝑓 ′ (𝑥) = lim 3 3 3 2 − ℎ ℎ→0 ℎ  = lim 3 CA ℎ→0 ℎ 2 − ℎ 3 𝑓′(𝑥) = lim ℎ→0 ℎ  answer/antwoord CA 2 (5) 𝑓′(𝑥) = − 3 6.2.1 𝑑𝑣 2 = 3 𝜋𝑟 2 ℎ 𝑑𝑟  answer/antwoord 4 = 𝜋𝑟ℎ CA 3 (1) 6.2.2 3 𝑥 2 −√3𝑥6 𝐷𝑥 [ √𝑥 4 + 4 𝑥3 4  𝑥 3 + 𝑥 −1 − √3𝑥 3 A 𝐷𝑥 [𝑥 3 + 𝑥 −1 − √3𝑥 3 ] 4 1  𝑥 3 CA 3 1 1 4 = 3 𝑥 − 𝑥 −2 − 3√3𝑥 2 3  − 𝑥2 CA 4 1 1  −3√3𝑥 2 CA = 3 𝑥 3 − 𝑥 2 − 3√3𝑥 2 (4) 6.3 𝑔(𝑥) = 3𝑥 2 + 𝑥 6.3.1 𝑔′(𝑥) = 6𝑥 + 1 𝑥=2  𝑔′(𝑥) = 6𝑥 + 1 A 𝑔′(𝑥) = 𝑚  substitution/substitusie 6(2) + 1 = 𝑚 CA 13 = 𝑚  answer/antwoord CA (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 11 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 6.3.2 𝑔(2) = 3(2)2 + 2  𝑦-value/waarde = 14 A 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )  substitution/substitusie 𝑦 − 14 = 13(𝑥 − 2) CA 𝑦 − 14 = 13𝑥 − 26  equation/vergelyking 𝑦 = 13𝑥 − 12 CA (3) 6.3.3 𝑔(1) = 3(1)2 + (1) = 4 (2; 14) and/en (1; 4) 𝑓(𝑏)−𝑓(𝑎) Average gradient/Gemiddelde gradiënt = 14−4 𝑏−𝑎  SF CA = 2−1  answer/antwoord = 10 CA (2) [18] QUESTION/VRAAG 7 7.1.1 𝑓(𝑥) = 𝑎(𝑥 − 𝑥1 )(𝑥 − 𝑥2 )(𝑥 − 𝑥3 )  SF A (−3; 0) (−1; 0) (2;0) 𝑓(𝑥) = 𝑎(𝑥 − (−3))(𝑥 − (−1))(𝑥 − 2)  simplification/ 𝑓(𝑥) = 𝑎(𝑥 + 3)(𝑥 2 − 𝑥 − 2) vereenvoudiging 𝑓(𝑥) = 𝑎(𝑥 3 + 2𝑥 2 − 5𝑥 − 6) CA (0; 6) 6 = 𝑎((0)3 + 2(0)2 − 5(0) − 6  −1 = 𝑎 CA 6 = −6𝑎 −1 = 𝑎  𝑏 = −2 CA 𝑓(𝑥) = −1(𝑥 3 + 2𝑥 2 − 5𝑥 − 6)  𝑐=5 CA 𝑓(𝑥) = −𝑥 3 − 2𝑥 2 + 5𝑥 + 6  𝑑=6 CA 𝑏 = −2 ; 𝑐 = 5 ; 𝑑 = 6 (6) 7.1.2 𝑓 ′ (𝑥) = −3𝑥 2 − 4𝑥 + 5  𝑓 ′ (𝑥) A 𝑓 ′ (𝑥) = 0  𝑓 ′ (𝑥) = 0 CA −3𝑥 2 − 4𝑥 + 5 = 0 from/vanaf 7.1.1  factors/faktore −(−4)±√(−4)2 −4(−3)(5) CA 𝑥= 2(−3)  both 𝑥-values/ 𝑥 = 0,79 or/of 𝑥 = −2,12 beide x-waardes CA 𝑓(−2,12) = −(−2,12)3 − 2(−2,12)2 + 5(−2,12) + 6 = −4,06 R(−2,12; −4,06) 𝑓(0,79) = −(0,79)3 − 2(0,79)2 + 5(0,79) + 6 = 8,21  R and/en T CA T(0,79; 8,21) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 12 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne 7.2.1 −3 ≤ 𝑥 ≤ −1 and/en 𝑥 ≥ 2  end points/ OR/OF eindpunte A 𝑥𝜖[−3; −1] and/en 𝑥𝜖[2; ∞)  notation/notasie  end points/ eindpunte A  notation/notasie A (4) 7.2.2 −2,12 < 𝑥 < 0,79  end points/ OR/OF eindpunte CA 𝑥𝜖(−2,12; 0,79) from/vanaf 7.1.2  notation/notasie CA (2) [17] QUESTION/VRAAG 8 8.1 960 𝑆. 𝐴(𝑥) = 4𝑥 + 𝑥 + 240  derivative/ afgeleide A 𝑆. 𝐴′ (𝑥) = 4 − 960𝑥 −2 ′ 960  𝑆. 𝐴 (𝑥) = 0 A 4 − 𝑥2 = 0  factors/faktore 4𝑥 2 = 960 CA 𝑥 2 = 240  𝑥-value/ x- 𝑥 = ±√240 waarde CA 𝑥 = ±15 Penalty: 1 mark for ∴ 𝑥 = 15 incorrect rounding/ Penaliseer: 1 punt vir foutiewe afronding (4) 8.2 960 𝑆. 𝐴(15) = 4(15) + (15) + 240  substitution/ substitusie CA = 364 𝑢𝑛𝑖𝑡𝑠/𝑒𝑒𝑛ℎ𝑒𝑑𝑒 2 from/vanaf 8.1  answer/antwoord CA NPR (2) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Technical Mathematics/P1/Tegniese Wiskunde/V1 13 NW/June/Junie 2025 Grade/Graad – 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 1 4𝑥 1 9.1.1 3  𝑥3 ∫ ( √𝑥 + (5) ) 𝑑𝑥 A 4 3 1 1 4𝑥  𝑥3 CA ∫ (𝑥 + (5) ) 𝑑𝑥 3 4 1 4𝑥 ( ) 1 4𝑥 3 4 ( ) 5  5 1 A = 4 𝑥3 + +𝐶 4𝑙𝑛 ( ) 1 5 4𝑙𝑛( ) 5  𝐶 A (4) 9.1.2 −3 ∫(𝑥 −1 − 𝑥) 𝑑𝑥  ∫(−3𝑥 −1 + 3𝑥)𝑑𝑥 ∫(−3𝑥 −1 + 3𝑥)𝑑𝑥  −3 ln 𝑥 A 3𝑥 2 3𝑥 2 = −3 ln 𝑥 + 2 + 𝐶  +𝐶 A 2 (3) 9.2 0 𝐴 = − ∫−2(3𝑥 3 − 12) 𝑑𝑥  area notation/ 0 oppervlakte 3𝑥 4 𝐴1 = − [ 4 − 12𝑥] notasie A −2  apply integration/ 3(0)4 3(−2)4 toepassing van 𝐴1 = − ( 4 − 12(0)) − ( − 12(−2)) integrasie A 4 𝐴1 = −(0 − 36)  substitution/ 𝐴1 = 36 𝑢𝑛𝑖𝑡𝑠 2 /𝑒𝑒𝑛ℎ𝑒𝑑𝑒 2 substitusie CA 2 𝐴2 = ∫0 (3𝑥 3 − 12) 𝑑𝑥  𝐴1 CA 2 3𝑥 4 𝐴2 = [ 4 − 12𝑥]  substitution/ 0 3(2)4 3(0)4 substitusie CA 𝐴2 = ( 4 − 12(2)) − ( 4 − 12(0)) 𝐴2 = (36 − 0)  𝐴2 CA 𝐴2 = 36 𝑢𝑛𝑖𝑡𝑠2 /𝑒𝑒𝑛ℎ𝑒𝑑𝑒2 𝐴𝑡𝑜𝑡𝑎𝑙 = 𝐴1 + 𝐴2 𝐴𝑡𝑜𝑡𝑎𝑙 = 36 + 36  𝐴𝑟𝑒𝑎𝑡𝑜𝑡𝑎𝑙 / 𝐴𝑡𝑜𝑡𝑎𝑙 = 72 𝑢𝑛𝑖𝑡𝑠/𝑒𝑒𝑛ℎ𝑒𝑑𝑒 2 𝑂𝑝𝑝𝑒𝑟𝑣𝑙𝑎𝑘𝑡𝑒𝑡𝑜𝑡𝑎𝑎𝑙 CA (7) [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou